Stuff

---Tonight: Lecture 4 July 23

---Assignment 2 has been posted.

---Presentation Assignment posted.

--Some more thermodynamics and then problem solving in class for Assignment #2.

--Next week: Free Energy and In-class problem solving session.

dU =�

∂U

∂T

�

V,ni

dT +�

∂U

∂V

�

T,ni

dV

∂qv = CvdT =�

∂U

∂T

�

V,n

First Law Differential Form

Total Differential

dU = CvdT +�

∂U

∂V

�

T,ni

dV Total Differential

1st Law of Thermodynamics

dU = ∂q − PextdV

Enthalpy Function: H(T,P)

dH =�

∂H

∂T

�

P

dT +�

∂H

∂P

�

T

dP

Total Differential

Large changes

At dP = 0 dH = ∂qp =�

∂H

∂T

�

P

dT

Cp =�

∂qp

∂T

�

p

=�

∂H

∂T

�

P

dH = CpdT +�

∂H

∂P

�

T,ni

dP Recasting

∆H = ∆U + ∆PV

dH = (∂q − PdV ) + PdV + V dP

dH = ∂q + V dP

Heat capacity of gases depends on P and V we define:

Heat Capacity Constant Volume

from large change ===> small change

Cp =�

∂qp

∂T

�

P,n

Cp =qp

∆T=

∆H

∆T

Cv =�

∂qv

∂T

�

V,n

Cv =qv

∆T=

∆U

∆T

Heat Capacity Constant Pressure

qp =� T2

T1

CpdTqv =� T2

T1

CvdT

KMT and the “equipartition of energy” links internal energy U and heat capacity, C of ideal gases.

Cv =�

∂qv

∂T

�

V

=

∂(

32nRT )

∂T

V

=32nR

U =32nRT

IMPORTANT: This equation says that the internal energy of a gas only depends on T and nothing else!

We can take the derivative and link the heat capacity to bulk properties

U =�

∂qv

∂T

�

V

n moles of monoatomic gas

U =32RT

U =32RT +

22RT + (3N − 5)RT

U =32RT +

22RT + (3N − 6)RT

(linear molecule)

(non-linear molecule)

(monoatomic)

Cv =32R +

22R + (3N − 6)R

Cv =32R +

22R + (3N − 5)R

Cv =32R

(linear molecule)

(non-linear molecule)

(monoatomic)

U

Cv

If we look at the results of the equipartion theorem we find the internal energies for linear and non-linear molecules.

3

3

Using the energies arising from equipartition of energy, show that the heat capacity at constant volume, Cv are the given values on the previous slides.

U =�

∂qv

∂T

�

V

U =32RT

U =32RT +

22RT + (3N − 5)RT

U =32RT +

22RT + (3N − 6)RT

(linear molecule)

(non-linear molecule)

(monoatomic)

3

The 1st Law tells us that energy is conserved, and sets up a formal accounting system to track energy transformation as heat and work.

---With calculus 1st law sets up experimental and theoretical parameters of interest to measure.

---But the 1st law does not give us predictive power on observation of naturally-occurring processes.

spontaneous notspontaneous

Spontaneity refers to a process that appears to proceed “naturally” from an initial state to a final state without outside intervention (does not mean “now”).

The rusting of a nail

Our goal in this Chapter is to identify those factors which determine whether a chemical reaction will proceed spontaneously.

Freezing and melting of water

T > 0˚C

T < 0˚C

In the 1870’s, it was thought that !H determined the “spontaneity of a chemical reaction”. This notion proved wrong.

Examples of spontaneous reactions:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) !H0 = -890.4 kJ

H2O (s) H2O (l) !H0 = 6.01 kJ

H2O (l) H2O (s) !H0 = - 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) !H0 = 25 kJH2O

!Hrnx does not determine whether a reaction is spontaneous!

T > 0C

T < 0C

What is the parameter or a bulk function that can predict whether A ==>B will occur “spontaneously” (irreversibly)?

2nd law of thermodynamics postulates that all processes that are spontaneous produce an increase in the entropy of the universe.

!Suniv = !Ssys + !Ssurr > 0

!Suniv = !Ssys + !Ssurr = 0

!Suniv = !Ssys + !Ssurr < 0 2. No spontaneous change

3. Equilibrium condition

1. Criteria for Spontaneous change!

There are many ways to express the 2nd Law of Thermodynamics

Deep: Work can be completely converted to heat, but heat can not be completely converted to work! (1st Law symmetry of work = heat is broken)

Kelvin Plank Statement: It is impossible for a system (engine) to undergo a cyclic process whose sole effects are flow of heat from a reservoir and the performance of an equal amount of work by the system.

Spontaneous processes increase the total entropy of the universe.

Practical: Heat can not spontaneously pass from a cold body to a hot body.

Entropy is too often described as positionally “disordered”. This is not correct, in the chemical sense, but it is useful device to predict.

How many ways can you have your room or a deck of cards organized vs disorganized?

0.5 atmevacuated1.0 atm

Gas expandsspontaneously

into larger volume

Energy level Energy level

Quantum mechancis dictates closer energy

level spacing as V increases.

Same amount of energy is dispersed or spread among more

energy levels.

It appears that entropy focuses on the most statistically favored position distribution entropy is more concerned with the fact that energy levels become more closely spaced and more occupied in that most favored state.

There is a thermodynamic state function called entropy, S, that is a measure of the energy dispersal that occurs when a change of state occurs.

-qsys reversible

T!S =

1. Bulk defintion (it needs no molecules).

!S = k (ln Wf - ln Wi)

Change in Entropy

2. Molecular Description

Macroscopic Microscopic

Entropy is a measure of the magnitude of energy dispersal over available quantum states in a chemical system. The more dispersed or spread out the energy is the higher the entropy.

Increasing the volume of a system decreases energy level spacing increasing entropy by populating more.

L 3L2LL

E1

E2

E3 E6

E5

E4

E3

E2

E1

E9

E8

E7

E6

E5

E4

E1

E2

E3

Boltzmann founded a thermodynamic state function called entropy, S, that is a measure of the “dispersion or spread of energy” that occurs when all spontaneous reactions occur.

When Wf > Wi then !S > 0

When Wf < Wi then !S < 0

!S = k ln Wf - k ln Wi

Change in Entropy

S = k ln WEntropy

# of microstates

Boltzman’s constant1.38 " 10#23 J/K.

Boltzmann’s Tomb In Vienna, Austria

A system can be described by bulk or macroscopic that we can observe and measure, and the microscopic or molecular that we can’t see but can model statistically. You must see both!

Macrostate is the observed state of the system that represents on a molecular level that microstate with the highest probability or number.

Microstate is particular distrubution that corresponds to some macrostate.

Multiplicity is the number of microstates that give a specific macrostate.

Consider 4 labeled molecules A,B,C,D

5-observable macrostates

A,B,CA,B,DA,C,DB,C,D

DCBA

A,B,C,D

Left Side

Right Side

Multiplicity

-

4

A,B C,DA,C B,DA,D B,CB,C A,DB,D A,CC,D A,B

6

1

A,B,C,D- 1

MicrostateBulk PropertyProbability

16

1/16

4/16

6/16

1/16

Macrostate Boltzman’s statistical interpretation of entropy (dispersal or spread of energy) is connected to a macroscopic defintion of heat flow per unit temperature.

-qsys reversible

T!S =

1. Bulk defintion (it needs no molecules).

!S = k (ln Wf - ln Wi)

Change in Entropy

2. Molecular DescriptionMacroscopic Microscopic

Increasing entropy or the dispersal or spread of energy occurs in all spontaneous processes. It provides humans with an “arrow of time” as the reverse situation never happens.

Thermal energy flows from the higher occupied energy levels in the warmer object into the unoccupied levels of the cooler one until equal numbers of states are occupied.

Example: two block of different temperatures are brought together.

100˚C 10˚C100˚C 10˚C

HOT COLD COMBINED

1 2

There are many chemical

reactions that lead to an

increase in dispersal of

energy spread over a larger

number microstates

entropy (!S > 0)

!S > 0

!S > 0

!S > 0

Dissolution

Dissolution

Mixing

Increasing T

Usual approach is Carnot Cycle

• Carnot analysis is long and the result unsatisfying

• Another approach is purely mathematical based on the Euler criteria and gives the same result.

• LET”S TRY IT

Can We Determine Mathematically Criteria For Spontaneous Change Using What We Know So Far?

KE gives this

KMT and the “equipartition of energy” links internal energy U and heat capacity, C of ideal gases.

Cv =�

∂qv

∂T

�

V

=

∂(

32nRT )

∂T

V

=32nR

U =32nRT IMPORTANT: This

equation says that the internal energy of a gas only depends on T and nothing else!

We can take the derivative and link the heat capacity to bulk properties

U =�

∂qv

∂T

�

V

n moles of monoatomic gas

Criteria For Spontaneous Change?

dU = ∂qrev + ∂wrev

rearrange

Assume monoatomic ideal gas U = 3/2 nRT

dU =�

∂U

∂T

�dT =

∂(

32nRT )

∂T

dT =32nRdT

from KMT

start with 1st law differential form

∂qrev =�

32nR

�dT +

�nRT

V

�dV

substituting

Is the above equation a total differential? How do we know or not know? Prove it is not.

∂qrev = dU + PextdV

dS =�

1T

�qrev =

�1T

� �32nR

�dT +

�1T

� �nRT

V

�dV

∂qrev =�

32nR

�dT +

�nRT

V

�dV

There is a math theorem that says we can transform an inexact differential to an exact one by multiplying by and integrating factor. Can we find one?

Show that this function is an exact differential using Euler’s criteria.

dS =∂qrev

T

∆S =�

dS =qrev

T

Only true for reversible path!

Does a change in entropy predict spontaneous processes? -Isolated 2-compartment system. Each

has its own Temp, Volume, U not at equilibrium.

No work, no energy or matter in or out, rigid container dV = 0.

Diathermal wall between compartments.

S = SA + SB

Our conditions impose the following:

dS =∂qrev

TdUA = ∂qrev = TAdSA

dUB = ∂qrev = TBdSB

UA + UB = constant ∴ UA = −UB

dS = dSA + dSB

!S > 0 predicts spontaneous processes.

dS =∂qrev

T

S = SA + SBUA + UB = constant ∴ UA = −UB

dS = dSA + dSB

=−dUB

TA+

dUB

TB

=dUA

TA+

dUB

TB

dS = dUB

�1

TB− 1

TA

�Analyze dS For:

Case 2: TA > TB Case 1: TB > TA

dS > 0dS increases for spontaneous flow of heatdS > 0

(-) (-) (+) (+)

A. Entropy Changes With Temperature @ Constant V

Problems Involving Entropy I

Nearly all problems start with the 1st Law or the enthalpy equations and involve restricting the path to eliminate variables.

dU = ∂q − PextdV

Case I. Entropy Changes With Temperature, dV = 0

dU = CvdT +�

∂U

∂V

�

S,ni

dV

dS =Cv

TdTdS =

∂qrev

T�

dS =� T2

T1

Cv(T )T

dT Is Cv constant?

B. Isothermal Expansion of An Ideal Gas

Problems Involving Entropy II

Nearly all problems start with the 1st Law or the enthalpy equations and involve restricting the path to eliminate variables.

dU = ∂q − PextdV

Case I. Entropy Changes ====> dT = 0, dU = 0

dU = CvdT +�

∂U

∂V

�

S,ni

dV

dU = ∂q − PextdV = 0

dS =P

TdV =

nR

VdV

∆S = nR ln�

V2

V1

�∆S = nR ln

�P2

P1

�

A. Isothermal Expansion of An Ideal Gas

Problems Involving Entropy III

Nearly all problems start with the 1st Law or the enthalpy equations and involve restricting the path to eliminate variables.

dU = ∂q − PextdV

Case I. Entropy Changes ====> dT = 0, dU = 0

dU = CvdT +�

∂U

∂V

�

S,ni

dV

dU = ∂q − PextdV = 0

dS =P

TdV =

nR

VdV

∆S = nR ln�

V2

V1

�∆S = nR ln

�P2

P1

�